# What are the absolute extrema of f(x)=(x-1)/(x^2+x+2) in[oo,oo]?

Oct 11, 2016

At $x = - 1$ the minimum
and at $x = 3$ the maximum.

#### Explanation:

$f \left(x\right) = \frac{x - 1}{{x}^{2} + x + 2}$ has stationary points characterized by

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{\left(x - 3\right) \left(1 + x\right)}{2 + x + {x}^{2}} ^ 2 = 0$ so they are at

$x = - 1$ and $x = 3$

Their characterization is made analyzing the signal of

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{2 \left(x \left(\left(x - 3\right) x - 9\right)\right) - 1}{2 + x + {x}^{2}} ^ 3$ at those points.

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(- 1\right) = 1 > 0 \to$ relative minimum
$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(3\right) = - \frac{1}{49} < 0 \to$ relative maximum.

Attached the function plot.